Is it true that every quasi-variety exends to a variety?

Question

I am wondering if the following is true:

If $Q$ is a quasivariety in an algebraic signature $S$, must there be a variety $V$ in an extended signature $S′⊇S$ such that $Q$ is the class of algebras isomorphic to $S$-reducts of algebras in $V$?

Answer 1

I will rephrase the question slightly, then give an affirmative answer.

Rephrased version: If $\mathcal Q$ is a quasivariety in an algebraic signature $S$, must there be a variety $\mathcal V$ in an extended signature $S'\supseteq S$ such that $\mathcal Q$ is the class of algebras isomorphic to $S$-reducts of algebras in $\mathcal V$?

Now for the affirmative answer to this version of the question.

The discriminator operation on a set $X$ is the function $d\colon X^3\to X$ defined by $ d(x,y,z) = z $ if $x=y$ and $d(x,y,z)=x$ if $x\neq y$. If $S$ is an algebraic signature and $\mathcal K$ is a class of algebras in a signature $S$, write $S'$ for the signature obtained by adding one ternary operation $d(x,y,z)$ to $S$, and write $\mathcal K^d$ to be the class obtained from $\mathcal K$ by interpreting $d$ as the discriminator operation on each member of $\mathcal K$.

Let $\mathcal Q$ be any $S$-quasivariety. Take $\mathcal K$ to be $\mathcal Q$, and take $\mathcal V$ to be the $S'$-variety generated by $\mathcal K^d$. It turns out that $\mathcal Q$ is the class of algebras isomorphic to $S$-reducts of algebras in $\mathcal V$. (See Theorem 9.4 of the book by Burris and Sankappanavar.)

Edit. Let me add more detail to this answer. Let $\mathcal K$ be any universal class. The class $\mathcal K^d$ (union the singleton $S$-algebra, if this class does not already contain the singleton $S$-algebra) is the class of algebras isomorphic to directly indecomposables in the variety ${\mathcal V}:=\textrm{HSP}({\mathcal K}^d)$, and every algebra in $\mathcal V$ is isomorphic to a Boolean product of members of ${\mathcal K}^d$. (This is easy to read off of Theorem 9.4, part (d).) But now, if $\mathcal K = \mathcal Q$ is a quasivariety, then it is a universal class containing a singleton algebra, so the variety $\mathcal V$ generated by ${\mathcal K}^d={\mathcal Q}^d$ is obtained from ${\mathcal Q}^d$ by closing under isomorphisms and Boolean products, i.e. ${\mathcal V}=I\Gamma^a({\mathcal Q}^d)$. The reducts of algebras in $\mathcal V$ to the signature $S$ will be the algebras in $I\Gamma^a({\mathcal Q}) = {\mathcal Q}$. Here I am using that quasivarieties are closed under isomorphisms and Boolean products. (Every Boolean product of some factors algebras is a subalgebra of a product of the factor algebras, and quasivarieties are closed under subalgebras and products.)